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Please show work. A. For single reactant: observe the change in rate for a chang

ID: 886401 • Letter: P

Question

Please show work.

A. For single reactant: observe the change in rate for a change in initial reactantconcentration.

The table above shows three independent reactions (not related to each other). What are the rate laws for each of thesereactions?

B. For multiple reactants: Find the reaction order for each reactant separately.

For the table above, find the rate law and the rate constant. What is the overall order of the reaction?

Reaction 1 Reaction 2 Reaction 3 [A] (M) Initial Rate (M/s) [C] (M) Initial Rate (M/s) [K] (M) Initial Rate (M/s) 0.10 0.015 0.10 0.015 0.10 0.015 0.20 0.030 0.20 0.060 0.30 0.135 0.40 0.060 0.40 0.240 0.90 1.215

Explanation / Answer

From the reaction we observe that there is only one reactant in each.

Rate law :

Reaction 1 ;

Rate = k [A]m

Here k is rate constant and m is order with respect to A.

We find m by using ratio of rate for two experiment

Lets take ratio of exp 2 to exp 1

Rate 2 / rate 1 = k [A]m/ k [A]m

Here k is constant and it is cancelled since reaction is same.

Lets plug concentration of A and rate value.

0.20 / 0.10 = [0.030]m/ [0.015]m

2 = (0.030/0.015)m

2 = 2m

To calculate m we need to take ln of both side

Ln 2 = m ln 2

m = 1

so rate law becomes

Rate = k [A]

Reaction : 2

Rate = k [C]n

We use same strategy to solve for n

Rate of Exp 2 to exp 1

Rate 2 / rate 1 = ( 0.060/0.015)n

0.2/0.1 = ( 0.060/0.015)n

2 = 4n

Ln 2 = n ln 4

n = ½

Rate law becomes

Rate = k [C]1/2

Reaction 3 :

Rate law :

Rate = k [K]l

Rate 2/ rate 1 = (0.135/0.015)l

0.30/0.10 =(0.135/0.015)l

3 = 9l

Ln 3 = l ln 9

l = 1/3

Rate law

Rate = k [K]1/3

B.

Rate law :

Rate = k [A]a [ X]b

Here a and b are the orders with respect to A and B respectively.

Ex. No.

[A] (M)

[X] (M)

Initial Rate (M/s)

1

0.050

0.050

0.014

2

0.150

0.050

0.056

3

0.050

0.100

0.028

4

0.150

0.150

0.168

Lets use ratio of rate 2/ rate 1

Rate 2/ Rate 1 = (A2/A1)a x (X2/X1)b

0.056 / 0.014 = (0.150/0.050)a   …concentration of X is same in both experiment so it is cancelled.

4 = 3a

We use ln

ln 4 = a ln 3

a = 1.26

Lets find for b

Rate 4/ Rate 2 = (A4/A2)a x (X4/X2)b

0.168/0.056 = (0.150 /0.050 )b

3 = 3b

b = 1

Rate law :

Rate = k [A]1.26 [X]

Calculation of rate constant

We use experiment first

0.014 = k [0.050]1.26 [0.050]

k = 12.27

Overall order : 1 + 1.26 = 2.26

Ex. No.

[A] (M)

[X] (M)

Initial Rate (M/s)

1

0.050

0.050

0.014

2

0.150

0.050

0.056

3

0.050

0.100

0.028

4

0.150

0.150

0.168